2. A lawn is in the form of a rectangle having its sides in the ratio 2: 3. The area of the lawn is (1/6) hectares. Find the length and breadth of the lawn.
Solution:
Let length = x and breadth = y. Then,
2 (x + y) = 46 or x + y = 23 and x2 + y2 = (17) 2 = 289.
Now, (x + y)2 = (23)2 <=> (x2 + y2) + 2xy = 529 <=> 289 + 2xy = 529 xy=120
Area = xy = 120 cm2
4. In measuring the sides of a rectangle, one side is taken 5% in excess, and the other 4% in deficit. Find the error percent in the area calculated from these measurements.
Solution:
Let x and y be the sides of the rectangle. Then,
Correct area = xy.
Calculated area = (105/100)*x * (96/100)*y = (504/500 )(xy)
Error In measurement = (504/500)xy- xy = (4/500)xy
Error % = [(4/500)xy *(1/xy) *100] % = (4/5) % = 0.8%. .
6.The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares.
Solution:
Side of first square = (40/4) = 10 cm.
Side of second square = (32/4)cm = 8 cm.
Area of third square = [(10)2 - (8)2] cm2 = (100 - 64) cm2 = 36 cm2.
Side of third square = (36)(1/2) cm = 6 cm.
Required perimeter = (6 x 4) cm = 24 cm.
Solution:
Ratio of their areas = (1/2)*(2x)2 :(1/2)*(5x)2 = 4x2 : 25x2 = 4 : 25.
9.The difference between two parallel sides of a trapezium is 4 cm. perpendicular distance between them is 19 cm. If the area of the trapezium is 475 find the lengths of the parallel sides.
Solution:
Let the two parallel sides of the trapezium be a em and b em.
Then, a - b = 4 And, (1/2) x (a + b) x 19 = 475
(a + b) =((475 x 2)/19)
a + b = 50
Solving (i) and (ii), we get: a = 27, b = 23.
So, the two parallel sides are 27 cm and 23 cm.
10. A person incures loss for by selling a watch for rs 1140.at what price should the watch be sold to earn a 5% profit ?
(brokerage (1/2) %)
Solution:
Let a = 13, b = 14 and c = 15. Then,
S = (1/2)(a + b + c) = 21.
(s - a) = 8, (s - b) = 7 and (s - c) = 6.
Area = (s(s - a) (s - b)(s - c))(1/2) = (21 *8 * 7*6)(1/2) = 84 cm2.
13.Find the area of a right-angled triangle whose base is 12 cm and hypotenuse is 13cm..
Solution:
Let the bases of the two triangles be x and y and their heights be 3h and 4h respectively.
Then, ((1/2) X x X 3h)/(1/2) X y X 4h) =4/3
x/y =(4/3 X 4/3)=16/9 Required ratio = 16 : 9.
15. Find the length of a rope by which a cow must be tethered in order tbat it may be able to graze an area of 9856 sq. metres.
Solution:
Clearly, the cow will graze a circular field of area 9856 sq. meters and radius equal to the length of the rope.
Let the length of the rope be R meters. Then,
π(R)^2 = (9856 X (7/22)) = 3136
R = 56.
Length of the rope = 56 m.
16.The area of a circular field is 13.86 hectares. Find the cost of fencing it at the rate of Rs. 4.40 per meter.
Solution:
(SP of 33m)-(CP of 33m)=Gain=SP of 11m
SP of 22m = CP of 33m
Let CP of each metre be Re.1 ,
Then, CP of 22m= Rs.22,SP of 22m=Rs.33.
Gain%=[(11/22)*100]%=50%
20.A vendor bought bananas at 6 for Rs.10 and sold them at Rs.4 for Rs.6 . Find his gain or loss percent.
